When 0.880 g of an organic compound containing carbon, hydrogen and oxygen is burned completely in oxygen, 1.760 g of co2 and 0.720 g h2 o are produced. what is the empirical formula of the compound?

1) Balance of mass

CxHyOz + O2 - - - > CO2 + H20

0.880 g + X = 1.760g + 0.720 g

=> X = 1.760g + 0.720g - 0.880g = 1.6 g of O2

2) C in CO2:

molar mass of CO2 = 44 g/mol

=> 12 g C / 44 g CO2 * 1.760 g CO2 = 0.480 g C

3) H in H2O

molar mass of H2O = 18.0 g/mol

2 g H / 18.0 g H2O * 0.720 g H2O = 0.080 g H

4) O

4a) O in CO2: 32gO / 44 g CO2 * 1.760 g CO2 = 1.280 g O

4b) O in H2O: 16gO / 18 gH2O * 0.720 g H2O = 0.640 g O

4c) O in CxHyOz = 1.280 g + 0.64 g - 1.6 g = 0.320 g O

5) Convert grams into moles

C: 0.480 g / 12 g/mol = 0.04 mol

H: 0.08 g / 1 g/mol = 0.08 mol

O: 0.32 g / 16 g/mol = 0.02 mol

6) Divide by the smallest number

C: 0.04 / 0.02 = 2

H: 0.08 / 0.02 = 4

O: 0.02 / 0.02 = 1

7) Use those numbers as subscripts for the empirical formula:

C2 H4 O

Answer: C2H4O